A magnifier for reading is in the form of a glass half...

A magnifier for reading is in the form of a glass half cylinder of radius $R$ and refractive index $n$. To use it, you place the flat side on the page you're trying to read, and several lines of text appear magnified. Find an expression for the magnification this device provides. Hint: With the object (the text) right at the surface of the magnifier, you can't treat the half cylinder as a thin lens. Instead, consult Fig. $31.36$, which shows an arrow as a small object being magnified. Assume the arrow's length $h$ is small enough that the horizontal distance from the arrowhead to the curved surface in Fig. $31.36$ is approximately the radius $R$, and that the angles $\alpha$ and $\gamma$ are therefore small enough that you can use the small-angle approxi sines and tangents.

Key Concepts

Refraction

Refraction is the phenomenon where light changes its direction as it passes from one medium to another due to a change in its speed. In optical devices like the half-cylinder magnifier, the refractive index governs how much the light bends at the curved interface, which is critical for determining how the image is formed and magnified.

Geometrical Optics

Geometrical optics uses ray diagrams and simple geometric principles to analyze the paths taken by light rays through optical systems. This approach is essential for understanding how the curved interface and the positioning of the object relative to it result in the magnified image observed when using the half-cylinder magnifier.

Magnification

Magnification in optics describes the ratio of the image size to the object size. It is derived by analyzing the ray paths and the corresponding angles created as light exits the optical device. For the half-cylinder magnifier, the overall geometry and the refractive properties of the material determine the degree of magnification experienced by the viewer.

Small-Angle Approximation

The small-angle approximation simplifies the analysis in optical systems by allowing the sine and tangent of small angles to be approximated by the angles themselves (when measured in radians). This approximation is particularly useful in this context, where the relevant angles are very small, enabling linear relationships between the angles and distances, which in turn facilitates the derivation of the magnification formula.

Transcript

00:02 In this problem we have a simple magnifier where the image is produced at a near point and we want to compute where the object is located.

00:19 So the object distance we call it p, so we want to compute the value of p and we know that the focal length of the lens is f.

00:33 Okay, so in order to compute the value of p we're going to use the value of p, we're going to use the value of p.

00:37 The lens equation that is this one and from here we can try to compute the value of p so 1 over p is 1 over f minus 1 over q and this cue is the image distance so the image distance is minus n and this is because n is the distance between the image and the image and the the lens, but the image distance is a virtual image and it is located on the same side that the object with respect of the lens.

01:31 So that's why this is negative.

01:34 It must be a negative.

01:36 So from here we have that 1 over b, it's just 1 over f minus 1 .1 .m.

01:46 Or is 1 over f plus 1 over n.

01:51 And from here, we can easily compute the value of p.

01:55 So p is just n times f over n plus f.

02:07 In the second part of the problem, we want to compute the value of the angular size.

02:15 So this angular size is this angle here.

02:19 And as this angle is very small, it is almost the same as its tangent.

02:31 So that's why theta is just h prime over n.

02:36 Also, we know that h prime over h is minus q over p.

02:45 This is because of the magnification of the lens.

02:50 And we already know that a q is minus minus n from here.

02:59 So we know that this is just n over p.

03:04 So from this last relation, we can get the value of h prime and we can say that this theta is just h prime and h prime is just this h that multiplies n over p.

03:20 So it's just n h, sorry h, and this over p...

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